Transcript 5 theoretical distributions

theoretical distributions
&
hypothesis testing
what is a distribution??
• describes the ‘shape’ of a batch of numbers
• the characteristics of a distribution can
sometimes be defined using a small number
of numeric descriptors called ‘parameters’
why??
• can serve as a basis for standardized
comparison of empirical distributions
• can help us estimate confidence intervals
for inferential statistics
• form a basis for more advanced statistical
methods
– ‘fit’ between observed distributions and certain
theoretical distributions is an assumption of
many statistical procedures
Normal (Gaussian) distribution
• continuous distribution
• tails stretch infinitely in both directions

180
168
156
144
132
120

108

96
84
72
60
48
36
24
12
0
1
2
3
4
5
6
7
8
9
10
11
12
• symmetric around the mean ()
• maximum height at 
• standard deviation () is at the point of
inflection
13
• a single normal curve exists for any
combination of , 
– these are the parameters of the distribution and
define it completely
• a family of bell-shaped curves can be
defined for the same combination of , ,
but only one is the normal curve
binomial distribution with p=q
• approximates a normal distribution of
probabilities
• p+q=1  p=q=.5
• =np=.5n
• recall that the binomial theorem specifies
that the mean number of successes is np;
substitute p by .5
0.300
• simplified from (n*0.25)
0.200
P(10,k,.5)
• =(np2)=.5n
0.250
0.150
0.100
0.050
0.000
0
2
4
6
k
8
10
• lots of natural phenomena in the real world
approximate normal distributions—near
enough that we can make use of it as a
model
• e.g. height
• phenomena that emerge from a large
number of uncorrelated, random events will
usually approximate a normal distribution
• standard probability intervals (proportions
under the curve) are defined by multiples of
the standard deviation around the mean
• true of all normal curves, no matter what 
or  happens to be
• P(- <=  <= +) = .683
• +/-1 = .683
• +/-2 = .955
• +/-3 = .997

180
168
156
144
132
120

108
• 50% = +/-0.67
• 95% = +/-1.96
• 99% = +/-2.58
96
84
72
60
48
36
24
12
0
1
2
3
4
5
6
7
8
9
10
11
12
13
• the logic works backwards
• if +/- < > .68, the distribution is not
normal
z-scores
• standardizing values by re-expressing them
in units of the standard deviation
• measured away from the mean (where the
mean is adjusted to equal 0)
xi  x
Zi 
s
• z-scores = “standard normal deviates”
• converting number sets from a normal
distribution to z-scores:
 presents data in a standard form that can be
easily compared to other distributions
 mean = 0
 standard deviation = 1
• z-scores often summarized in table form as
a CDF (cumulative density function)
• Shennan, Table C (note errors!)
• can use in various ways, including
determining how different proportions of a
batch are distributed “under the curve”
Neanderthal stature
• population of Neanderthal skeletons
• stature estimates appear to follow an
approximately normal distribution…
– mean = 163.7 cm
– sd = 5.79 cm
Quest. 1: what proportion of the
population is >165 cm?
• z-score = ?
• z-score = (165-163.7)/5.79 = .23 (+)
mean = 163.7 cm
sd = 5.79 cm
.48803 .48405 .48006 .47608
Quest. 1: what proportion of the
population is >165 cm?
• z-score = .23 (+)
• using Table C-2
– cdf(.23) = .40905
– 40.9%
Quest. 2: 98% of the population
fall below what height?
• Cdf(x)=.98
• can use either table
– Table C-1; look for .98
– Table C-2; look for .02
.48803 .48405 .48006 .47608
Quest. 2: 98% of the population
fall below what height?
• Cdf(x)=.98
• can use either table
– Table C-1; look for .98
– Table C-2; look for .02
– both give you a value of 2.05 for z
• solve z-score formula for x: xi
• x = 2.05*5.79+163.7 = 175.6cm
 Z i  x
“sample distribution of the mean”
• we don’t know the shape of the distribution
an underlying population
• it may not be normal
• we can still make use of some properties of
the normal distribution
• envision the distribution of means associated
with a large number of samples…
central limits theorem
• distribution of means derived from sets of
random samples taken from any population
will tend toward normality
• conformity to a normal distribution
increases with the size of samples
• these means will be distributed around the
mean of the population
Xx  
• we usually have one of these samples…
• we can’t know where it falls relative to the
population mean, but we can estimate odds
about how far it is likely to be…
• this depends on
– sample size
– an estimate of the population variance
• the smaller the sample and the more
dispersed the population, the more likely
that our sample is far from the population
mean
• this is reflected in the equation used to
calculate the variance of sample means:
s 
2
x

2
n
• the standard deviation of sample means is the
standard error of the estimate of the mean:
se 

1



n
n
n
2
• you can use the standard error to calculate
a range that contains the population mean,
at a particular probability, and based on a
specific sample:
x  Z
s
n
(where Z might be 1.96 for .95 probability, for example)
ex. Shennan (p. 81-82)
• 50 arrow points
– mean length = 22.6 mm
– sd = 4.2 mm
•
•
•
•

4.2
s 

 .594
n
50
standard error = ??
22.6 +/- 1.96*.594
22.6 +/- 1.16
95% probability that the population mean is
within the range 21.4 to 23.8
hypothesis testing
• originally used where decisions had to be
made
• now more widely used—even where
evaluation of data would be more
appropriate
• involves testing the relative strength of null
vs. alternative hypotheses
“null hypothesis”
H0
• usually highly specific and explicit
• often a hypothesis that we suspect is
wrong, and wish to disprove
• e.g.:
1. the means of two populations are the same
(H0:1=2 )
2. two variables are independent
3. two distributions are the same
“alternative hypothesis”
H1
• what is logically implied when H0 is false
• often quite general or nebulous compared to
H0
• the means of two populations are different:
H1:1< >2
testing H0 and H1
• together, constitute mutually exclusive and
exhaustive possibilities
• you can calculate conditional probabilities
associated with sample data, based on the
assumption that H0 is correct
• P(sample data|H0 is correct)
• if the data seem highly improbable given
H0, H0 is rejected, and H1 is accepted
• what can go wrong???
• since we can never know the true state of
underlying population, we always run the
risk of making the wrong decision…
Type 1 error
• P(rejecting H0|H0 is true)
• probability of rejecting a true null
hypothesis
– e.g.: deciding that two population means are
different when they really are the same
• P = significance level of the test = alpha ()
• in “classic” usage, set before the test
• smaller alpha values are more conservative
from the point of view of Type I errors
• compare a alpha-level of .01 and .05:
– we accept the null hypothesis unless the sample
is so unusual that we would only expect to
observe it 1 in 100 and 5 in 100 times
(respectively) due to random chance
– the larger value (.05) means we will accept less
unusual sample data as evidence that H0 is false
– the probability of falsely rejecting it
(i.e., a Type I error) is higher
• the more conservative (smaller) alpha is set
to, the greater the probability associated
with another kind of error—Type II error
Type II error
• P(accepting H0|H0 is false)
• failing to reject the null hypothesis when it
actually is false
• the probability of a Type II error () is
generally unknown
• the relative costs of Type I vs. Type II errors
vary according to context
• in general, Type I errors are more of a problem
• e.g., claiming a significant pattern where none
exists
H0 is correct
H0 is incorrect
H0 is accepted
correct decision
Type II error ()
H0 is rejected
Type I error ()
correct decision
example 1
• mortuary data (Shennan, p. 56+)
• burials characterized according to 2 wealth
(poor vs. wealthy) and 6 age categories
(infant to old age)
Rich
Poor
Infans I
6
23
Infans II
8
21
Juvenilis
11
25
Adultus
29
36
Maturus
19
27
Senilis
3
4
Total
76
136
• counts of burials for the younger ageclasses appear to be disproportionally high
among “poor” burials
• can this be explained away as an example of
random chance?
or
• do poor burials constitute a different
population, with respect to age-classes, than
rich burials?
• we might want to make a decision about
this…
• we can get a visual sense of the problem
using a cumulative frequency plot:
1
0.9
0.8
rich
0.7
poor
0.6
0.5
0.4
0.3
0.2
0.1
Senilis
Maturus
Adultus
Juvenilis
Infans II
Infans I
0
• K-S test (Kolmogorov-Smirnov test) assesses the
significance of the maximum divergence between two
cumulative frequency curves
H0:dist1=dist2
• an equation based on the theoretical distribution of
differences between cumulative frequency curves
provides a critical value for a specific alpha level
• differences beyond this value can be regarded as
significant (at that alpha level), and not attributed to
random processes…
• if alpha = .05, the critical value =
1.36*(n1+n2)/n1n2
1.36*(76+136)/76*136 = 0.195
1
0.8
rich
0.7
poor
0.6
Dmax=.178
0.5
0.4
0.3
0.2
0.1
Senilis
Maturus
Adultus
Juvenilis
Infans II
0
Infans I
• the observed value = 0.178
• 0.178 < 0.195; don’t reject H0
• Shennan: failing to reject H0 means
there is insufficient evidence to
suggest that the distributions are
different—not that they are the
same
• does this make sense?
0.9
example 2
• survey data  100 sites
• broken down by location and time:
early
late
Total
piedmont
31
19
50
plain
19
31
50
Total
50
50
100
• we can do a chi-square test of independence
of the two variables time and location
• H0:time & location are independent
• alpha = .05
time
location
time
location
H0
H1
• 2 values reflect accumulated differences between
observed and expected cell-counts
• expected cell counts are based on the assumptions
inherent in the null hypothesis
• if the H0 is correct, cell values should reflect an
“even” distribution of marginal totals
piedmont
plain
Total
early
25
late
50
50
Total
50
50
100
• chi-square = ((o-e)2/e)
• observed chi-square = 4.84
• we need to compare it to the “critical value”
in a chi-square table:
• chi-square = ((o-e)2/e)
• observed chi-square = 4.84
• chi-square table:
 critical value (alpha = .05, 1 df) is 3.84
 observed chi-square (4.84) > 3.84
• we can reject H0
• H1: time & location are not independent
• what does this mean?
early
late
Total
piedmont
31
19
50
plain
19
31
50
Total
50
50
100
example 3
• hypothesis testing using binomial
probabilities
• coin testing: H0:p=.5
• i.e. is it a fair coin??
• how could we test this hypothesis??
• you could flip the coin 7 times, recording
how many times you get a head
• calculate expected results using binomial
theorem for P(7,k,.5)
k
0
1
2
3
4
5
6
7
p
0.5
P(7,k,.5)
0.008
0.055
0.164
0.273
0.273
0.164
0.055
0.008
0.300
0.250
P(7,k,.5)
n
7
0.200
0.150
0.100
0.050
0.000
0
1
2
3
4
k
5
6
7
• define rejection subset for some level of alpha
• it is easier and more meaningful to adopt nonstandard  levels based on a specific rejection set
• ex:
{0,7}
 = .016
n
7
P(7,k,.5)
0.300
0.250
0.200
0.150
0.100
0.050
0.000
0
1
2
3
4
k
5
6
7
k
0
1
2
3
4
5
6
7
p
0.5
P(7,k,.5)
0.008
0.055
0.164
0.273
0.273
0.164
0.055
0.008
{0,7}; =.016
• under these set-up conditions, you reject H0 only if
you get 0 or 7 heads
• if you get 6 heads, you accept the H0 at a alpha
level of .016 (1.6%)
• this means that IF THE COIN IS FAIR, the
outcome of the experiment could occur around 1
or 2 times in 100
• if you have proceeded with an alpha of .016, this
implies that you regard 6 heads as fairly likely
even if H0 is correct
• but you don’t really want to know this…
• what you really want to know is
IS THE COIN FAIR??
• you may NOT say that you are 98.4% sure
that the H0 is correct
– these numerical values arise from the
assumption that H0 IS correct
– but you haven’t really tested this directly…
{0,1,6,7}; =.126
• you could increase alpha by widening the
rejection set
• this increases the chance of a Type I error—
doubles the number of outcomes that could
lead you to reject the null hypothesis
• it makes little sense to set alpha at .05
• your choices are really between .016 and .126
problems…
a) hypothesis testing often doesn’t answer
very directly the questions we are interested
in
– we don’t usually have to make a decision in
archaeology
– we often want to evaluate the strength or
weakness of some proposition or hypothesis
• we would like to use sample data to tell us
about populations of interest:
P(P|D)
• but, hypothesis testing uses assumptions
about populations to tell us about our
sample data:
P(D|P) or P(D|H0 is true)
b) classical hypothesis testing encourages
uncritical adherence to traditional
procedures
“fix the alpha level before the test, and never
change it”
“use ‘standard’ alpha levels: .05, .01”
 if you fail to reject the H0, there seems to
be nothing more to say about the matter…
early
late
Total
piedmont
31
19
50
plain
19
31
50
Total
50
50
100
early
late
Total
piedmont
29
20
49
plain
21
30
51
Total
50
50
100
(shift 3 sites)
no longer
significant at
alpha = .05 !
early
late
Total
piedmont
31
19
50
plain
19
31
50
Total
50
50
100
early
late
Total
piedmont
29
20
49
plain
21
30
51
Total
50
50
100
 = .016
 = .072
• better to report the actual alpha value
associated with the statistic, rather than just
whether or not the statistic falls into an
arbitrarly defined critical region
• most computer programs do return a
specific alpha level
• you may get a reported alpha of .000
• not the same as “0”
• means  < .0005 (report it like this)

2
critical:
observed:

.05
accept H0
reject H0
3.84
4.84
.016
2
observed:
4.84
c) encourages misinterpretation of results
• it’s tempting (but wrong) to reverse the
logic of the test
– having failed to reject the H0 at an alpha of .05,
we are not 95% sure that the H0 is correct
– if you do reject the H0, you can’t attach any
specific probability to your acceptance of H1
d) the whole approach may be logically
flawed:
• what if the tests lead you to reject H0?
• this implies that H0 is false
• but the probabilities that you used to reject it are
based on the assumption that H0 is true; if H0 is
false, these odds no longer apply
• rejecting H0 creates a catch-22; we accept the H1,
but now the probabilistic evidence for doing so is
logically invalidated
Estimation
• [revisit later, if time permits…]